Question:

The equilibrium constant, \(K_C\) for \(3C_2H_2(g) \rightleftharpoons C_6H_6(g)\) is 4 L\(^2\)mol\(^{-2}\). If the equilibrium concentration of benzene is 0.5 mol/L, then what is the value of concentration of ethylene?

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Remember to cube the concentration of the reactant when using equilibrium constant expressions for reactions involving stoichiometric coefficients.
Updated On: Apr 16, 2025
  • 0.05 mol/L
  • 0.5 mol/L
  • 0.25 mol/L
  • 0.025 mol/L
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The Correct Option is B

Solution and Explanation


The balanced equation is: \[ 3C_2H_2(g) \rightleftharpoons C_6H_6(g) \] Let the concentration of ethylene \(C_2H_2\) at equilibrium be \(x\) mol/L, and the concentration of benzene \(C_6H_6\) is given as 0.5 mol/L. Using the equilibrium constant expression: \[ K_C = \frac{[C_6H_6]}{[C_2H_2]^3} \] Substitute the given values: \[ 4 = \frac{0.5}{x^3} \] Now, solve for \(x\): \[ x^3 = \frac{0.5}{4} = 0.125 \] \[ x = \sqrt[3]{0.125} = 0.5 \] Thus, the concentration of ethylene is 0.5 mol/L.
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